Previously we observed that although monomorphisms tended to give expected generalizations of injective function in many categories, epimorphisms sometimes weren’t the expected generalization of surjective functions. We also discussed split epimorphisms, but where the definition of an epimorphism is too permissive to agree with the surjective morphisms in familiar concrete categories, the definition of a split epimorphism is too restrictive.

In this post we will discuss two other intermediate notions of epimorphism. (These all give dual notions of monomorphisms, but their epimorphic variants are more interesting as a possible solution to the above problem.) There are yet others, but these two appear to be the most relevant in the context of abelian categories.

**Regular**

Recall that the **equalizer** of a parallel pair of morphisms is their limit. More explicitly, it is the universal object together with a map such that . A map with this property is said to **equalize** and , so the equalizer is the universal equalizing map.

Dually, the **coequalizer** of a parallel pair of morphisms is their colimit. More explicitly, it is the universal object together with a map such that .

The definition of equalizer and coequalizer can in principle be extended to any family of morphisms, but generally speaking the terms “equalizer” and “coequalizer” refer to a pair of morphisms.

*Example.* In , the equalizer of a parallel pair of functions is , and the coequalizer of is the quotient of by the identifications . Roughly speaking, equalizers cut out subsets of the source where some condition holds while coequalizers force certain conditions to hold in the target.

*Sub-example.* If is an endomorphism, the equalizer of is the set of fixed points of .

*Example.* Let be any concrete category whose forgetful functor has a left adjoint. Then preserves limits, so if equalizers exist in , then on underlying sets they look like they do above. Coequalizers may look different; for example, in , the coequalizer of a parallel pair of morphisms is the quotient of by the normal subgroup generated by the elements .

*Example.* The forgetful functor has both a left and right adjoint, so preserves both equalizers and coequalizers. Consequently the above description carries over to as well. The coequalizer in particular should be thought of as a mild generalization of the notion of a quotient space.

*Example.* Let be a group acting on an object in a category . Then the equalizer of all of the morphisms describes the -invariants of , while their coequalizer describes the -coinvariants of .

*Sub-example.* If and is an abelian group, then and .

*Example.* In any -enriched category, the equalizer of a parallel pair of morphisms is the equalizer of , otherwise known as the **kernel** of . In this case we write and .

Dually, the coequalizer of a parallel pair of morphisms is the coequalizer of , otherwise known as the **cokernel** of . In this case we write and .

The kernel and cokernel as defined here reproduce the usual notion of kernel and cokernel in, say, for a ring. The kernel and cokernel make sense in any category with zero morphisms, but computing kernels and cokernels is not in general equivalent to computing equalizers and coequalizers in such a category since we may not have the ability to subtract morphisms.

**Definition-Proposition:** The equalizer of some pair of morphisms is a **regular monomorphism**. In particular, a regular monomorphism is a monomorphism.

*Proof.* Let be a pair of morphisms such that . Call this composite . Then equalizes , hence factors uniquely through . Since both constitute factorizations of through , it follows that .

Dually, we obtain the following.

**Definition-Proposition:** The coequalizer of some pair of morphisms is a **regular epimorphism**. In particular, a regular epimorphism is an epimorphism.

Note that, like being a monomorphism (resp. epimorphism), being a regular monomorphism (resp. regular epimorphism) is a limit property (resp. colimit property).

Regular epimorphisms should be thought of as being generalized quotient maps in a sense which will be made precise in the next section. We defer examples until the next section, where we will prove a result which will allow us to verify examples in many classes of categories. For now, consider the following.

**Proposition:** Let be a parallel pair of morphisms and let be a monomorphism. Then the induced map is an isomorphism.

*Proof.* It suffices to verify that the two equalizers have the same universal property: a map satisfies if and only if it satisfies since is left-cancellative.

Dually, we obtain the following.

**Proposition:** Let be a parallel pair of morphisms and let be an epimorphism. Then the induced map is an isomorphism.

Roughly speaking, the above propositions say that equalizers don’t change if we embed the target into something bigger and coequalizers don’t change if we think of the source as the quotient of something bigger.

**Effective**

The definition of a regular monomorphism (resp. epimorphism) only refers to being the coequalizer of some pair of maps. However, in order to determine whether a morphism is a regular monomorphism (resp. epimorphism), there is a natural pair of maps to consider if it exists.

Let be a morphism in a category. The **kernel pair** of is the pullback of the diagram ; equivalently, it is the fiber product . Dually, the **cokernel pair** of is the pushout of the diagram ; equivalently, it is the amalgamated coproduct .

The following examples will only discuss kernel pairs. We already saw some examples of cokernel pairs earlier when trying to determine whether various maps were epimorphisms. Kernel pairs should be thought of as internal equivalence relations or **congruences** associated to a map; this can be given a formal definition but we will not need it.

*Example.* Let be a map of sets. Then

.

In other words, as a subset of , it is precisely the equivalence relation induced by on .

Note that the above subset or subspace does not change upon replacing the codomain by ; in other words, to compute kernel pairs we may assume without loss of generality that is surjective. This generalizes.

Note also that if is surjective then the coequalizer of the two projections is . This also generalizes.

*Example.* Let be any concrete category whose forgetful functor has a left adjoint. Then preserves limits, so if kernel pairs exist in , then on underlying sets they look like they do above. This is true in particular for . Here an alternate interpretation of the kernel pair of a group homomorphism with kernel is that it consists of the subgroup of pairs such that ; similarly, if is a ring homomorphism with kernel , then the kernel pair consists of the subgroup of pairs such that .

Recall that in there is a natural way to pass back and forth between equivalence relations on a set and surjective functions with source . One direction of this equivalence is given by taking the kernel pair of . This produces a subset of together with a pair of projections . The other direction of this equivalence is given by taking the coequalizer of . This suggests the following definition.

**Definition:** A morphism is an **effective epimorphism** if it has a kernel pair and is the coequalizer of its kernel pair.

(Note that the data of the kernel pair includes the data of a pair of maps to , and as above it is this pair of maps that we take the coequalizer of.)

Dually, we obtain the following definition.

**Definition:** A morphism is an **effective monomorphism** if it has a cokernel pair and is the equalizer of its cokernel pair.

Note that an effective epimorphism (resp. monomorphism) is in particular a regular epimorphism (resp. monomorphism). In many categories of interest, the converse is also true.

**Proposition:** Let be a morphism with a kernel pair . Then is a regular epimorphism if and only if it is an effective epimorphism.

*Proof.* One direction is clear. In the other, suppose is the coequalizer of some parallel pair of morphisms . By assumption, has a kernel pair, so factor through it: that is, there is a map such that, letting be the two projections, we have .

It remains to show that is the coequalizer of . But if is another map such that , then

so also coequalizes , hence factors through by assumption.

Dually, we obtain the following.

**Proposition:** Let be a morphism with a cokernel pair . Then is a regular monomorphism if and only if it is an effective monomorphism.

In particular, in a category with finite limits and colimits (which include all of the categories we discuss below), regular and effective coincide.

*Example.* The effective epimorphisms in are precisely the surjective functions. This is just the statement that a surjective function is the quotient of by the equivalence relation on it induces. The effective monomorphisms are also precisely the injective functions. This follows from an explicit description of the cokernel pair of an injective function , which is given by the disjoint union of together with two copies of .

*Example.* The forgetful functor preserves both limits and colimits, hence preserves kernel pairs, cokernel pairs, equalizers, and coequalizers. Equalizers are equipped with subspace topologies and coequalizers are equipped with quotient topologies. Consequently, the effective epimorphisms are precisely the quotient maps, and the effective monomorphisms are precisely the embeddings.

*Example.* The forgetful functor preserves limits but not colimits, hence preserves kernel pairs and equalizers but not cokernel pairs or coequalizers. However, it is still true that the effective epimorphisms are precisely the surjective homomorphisms; this is a form of the first isomorphism theorem. It also follows from the nontrivial fact we used about amalgamated free products earlier that the effective monomorphisms are precisely the injective homomorphisms.

*Example.* As above, the effective epimorphisms in are precisely the surjective homomorphisms, and as above, this is a form of the first isomorphism theorem (for rings). So we have successfully given a categorical definition of a surjective homomorphism of rings.

Effective monomorphisms do not behave as well. Recall that if is a commutative ring, then any nontrivial localization is an epimorphism, hence has cokernel pair . It follows that the equalizer of the cokernel pair of is the identity map , hence is not , so is not an effective monomorphism although it is a monomorphism.

**Isomorphisms**

Previously we saw that an isomorphism is equivalently either a morphism which is both a monomorphism and a split epimorphism or a morphism which is both a split monomorphism and an epimorphism. This statement can be strengthened.

**Theorem:** The following conditions on a morphism are equivalent:

- is an isomorphism.
- is a monomorphism and a regular epimorphism.
- is a regular monomorphism and an epimorphism.

*Proof.* An isomorphism is both an equalizer and a coequalizer of any parallel pair of identical morphisms, so is in particular both a regular epimorphism and a regular monomorphism. Since being an isomorphism is a self-dual condition, it suffices to show that if is a monomorphism and a regular epimorphism then it is an isomorphism.

So suppose that is the coequalizer of a pair of morphisms. Then in particular , but by left cancellability . The coequalizer of any pair of identical morphisms is necessarily an isomorphism (since the identity map needs to factor through it and vice versa), so the conclusion follows.

**Corollary:** Suppose is an epimorphism but not an isomorphism. Then is not a regular monomorphism.

This generalizes the observation we made earlier about effective monomorphisms in .

Strictly speaking, the above theorem does not imply the previous theorem until we know the following.

**Proposition:** A split epimorphism is regular.

*Proof.* Let be a split epimorphism, hence there exists a section (so ). The composite is an idempotent. We claim that

.

To see this, let be a morphism such that . Then , hence factors through as desired.

Dually, we obtain the following.

**Proposition:** A split monomorphism is regular.

However, we obtain more than this: the proof above also shows that

hence both and can be recovered from the idempotent acting on alone. An idempotent such that the coequalizer and equalizer above both exist (equivalently, such that there exist such that ) is said to be **split**.

Note that and above are coequalizers resp. equalizers for purely diagrammatic reasons, so this remains true after an arbitrary functor has been applied; that is, the above are absolute coequalizers resp. equalizers.

on November 4, 2012 at 12:24 am |Zhen LinThere’s an interesting corollary to the proposition that a morphism is a regular epimorphism if and only if it is the coequaliser of its kernel pair: assuming the axiom of choice, epimorphisms in Set are preserved by any functor whatsoever; this can then be used to show that the forgetful functor from the Eilenbergâ€“Moore category of a monad over Set preserves coequalisers _of kernel pairs_, and hence, the _regular_ epimorphisms in the Eilenbergâ€“Moore category are exactly those whose underlying map is a surjection.

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